tìm a, b, c thuộc N* biết\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{4}{5}\)
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Cho mình sửa lại đề câu 1b: \(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\)
\(\frac{x}{7}-\frac{1}{2}=\frac{1}{y+1}\)
\(\frac{2x-7}{14}=\frac{1}{y+1}\)
\(TH1:\hept{\begin{cases}2x-7=7\\y+1=2\end{cases}\Rightarrow\hept{\begin{cases}x=7\\y=1\end{cases}}}\)
\(TH2:\hept{\begin{cases}2x-7=-7\\y+1=-2\end{cases}\Rightarrow\hept{\begin{cases}x=0\\y=-3\end{cases}}}\)
nhớ cho
Ta có : \(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)
Vậy a = 1,b = 2,c = 3,d = 4
\(\frac{30}{43}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{\frac{43}{30}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)
\(\Rightarrow\frac{1}{1+\frac{13}{30}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{1+\frac{1}{2+\frac{1}{\frac{13}{4}}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)
\(\Rightarrow a=1,b=2,c=3,d=4\)
Ta có: \(\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}=\frac{1}{1+\frac{1}{2+\frac{1}{\frac{13}{4}}}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{\frac{43}{30}}=\frac{30}{43}\)
Vậy a=1
b=2
c=3
d=4
\(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)
a=1;b=2;c=3;d=4
\(\frac{1.bc}{abc}+\frac{1.ac}{abc}+\frac{1.ab}{abc}=1\)
\(bc+ac+ab=abc\)
phần sau bạn làm nốt nhé